Răspuns:
0
Explicație pas cu pas:
[tex]S = {1}^{2019} + {2}^{2019} + {3}^{2019} + ... + {2019}^{2019} \\ [/tex]
[tex](a + b)|({a}^{n} + {b}^{n}), \: n \: impar[/tex]
[tex]a ≡ -b \: (mod \: a + b) => {a}^{n} ≡ {( - b)}^{n} (mod \: a + b) \\ [/tex]
[tex]{a}^{n} ≡ -{b}^{n} \: (mod \: a + b), \: n \: impar[/tex]
[tex]=> {a}^{n} + {b}^{n} ≡ 0 \: (mod \: a + b)[/tex]
[tex]{1}^{2019} + {29}^{2019} ≡ 0 \: mod \: (1 + 29) ≡ 0 \: mod \: 30 \\ [/tex]
[tex]{2}^{2019} + {28}^{2019} ≡ 0 \: mod \: 30 \\ [/tex]
...
[tex]{14}^{2019} + {16}^{2019} ≡ 0 \: mod \: 30 \\ [/tex]
[tex]{15}^{2019} ≡ 15 \: mod \: 30 \\ [/tex]
[tex]{30}^{2019} ≡ 0 \: mod \: 30 \\ [/tex]
Astfel:
[tex]{1}^{2019} + {2}^{2019} + {3}^{2019} + ... + {30}^{2019} ≡ 15 \: mod \: 30 \\ [/tex]
2010 = 67×30, deci:
[tex]{1}^{2019} + {2}^{2019} + ... + {2010}^{2019} ≡ 15 \: mod \: 30 \\[/tex]
iar:
[tex]{2011}^{2019} + {2012}^{2019} + ... + {2019}^{2019} ≡ \\ ≡ (1 + 8 + 27 + 4 + 5 + 6 + 13 + 2 + 9) \: mod \: 30 \\ ≡ 75 \: mod \: 30 ≡ 15 \: mod \: 30 \\[/tex]
în final:
[tex]{1}^{2019} + {2}^{2019} + {3}^{2019} + ... + {2019}^{2019} ≡ 0 \: mod \: 30 \\[/tex]
